Question

已知不等式ax/(x-1)＜1的解集为{x|x＜1或x＞2}，则a=？

Answer 1

解ax/(x-1)＜1
等价于ax/(x-1)-1＜0
即ax/(x-1)-(x-1)/(x-1)＜0
即[ax-x+1]/(x-1)＜0
等价于[ax-x+1]*(x-1)＜0
即不等式ax/(x-1)＜1的解集为{x|x＜1或x＞2}，
为不等式[ax-x+1]*(x-1)＜0的解集为{x|x＜1或x＞2}，
即[ax-x+1]*(x-1)=0的根x=1或x=2
把x=2代入[ax-x+1]*(x-1)=0
即[a*2-2+1]*(2-1)=0
即a*2-2+1=0
即2*a=1
即a=1/2

Answer 2

ax/(x-1)<1
ax/(x-1) -1<0
(ax -x+1)/(x-1)<0
[(a-1)x +1]/(x-1)<0
[(1-a)x -1]/(x-1)>0
由解集构造不等式
(x-2)/(x-1)>0
[(1/2)x -1]/(x-1)>0
两不等式为同一不等式
1-a =1/2
a=1/2

Answer 3

ax/(x-1)＜1
ax/(x-1)- 1＜0
【（a-1）x+1】/（x-1）＜0
等价于【（a-1）x+1】（x-1）＜0
解集为{x|x＜1或x＞2}，
那么（a-1）x+1=0的解为2
2（a-1）+1=0
a=1/2